hello : x = 4 + ln t....(1) y = t² + 3....(2) 1) the first method : by (1) : lnt = x-4 t = e^(x-4) subsct in (2) : y = (e^(x-4))²+3 y = e^(2x-8)+3 let : f(x) = e^(2x-8)+3 an equation of the tangent at (4,4) is : y = f'(4)(x-4) +f(4) f'(x) = 2e^(2x-8) f'(4) =2e^(0) =2 f(4) = 4 so : y = 2(x-4)+4 y= 2x-4
